# 3+1 slicing of spacetime question

Gold Member
This has to do with the ADM formulation of GR. I am following MTW chapter 21 and Wald appendix E and chapter 10.

On page 510 in MTW, they are talking about the covariant derivative on the hypersurface defined via the covariant derivative on the 4-manifold. They take the 4-D covariant derivative of an arbitrary vector in equation 21.54 and then go on to comment on the fact that it has a component proportional to $\mathbf{e}_0\cdot \mathbf{n}$ which does not lie on the hypersurface, and they claim that to take the 3-D covariant derivative is to project away (kill in their language) this component so that we are left with a vector which lies entirely in the hypersurface. I understand them up until here.

What I don't understand is their next statement that says one can do this by simply changing the mu index which runs from 0 to 3 into an m index which runs from 1 to 3 in equation 21.54. Since the term which actually appears in 21.54 is a term that is proportional to $\mathbf{e}_0$, we should ONLY be able to project away the normal component of this. Given that $\mathbf{e}_0=N\mathbf{n}+\mathbf{N}$, we should be left with a term that is proportional to $\mathbf{N}$ shouldn't we? I don't see how we can just get rid of that term.

Later on in the same section, MTW discusses the covariant derivative of a oneform. In there, he seems to insist on using only the Christoffel symbols of the first kind (all lower indices), is there a reason for this? It seems natural to use the regular Christoffel symbols of the second kind (one upper and 2 lower indices)...it just seems a little weird that he would do this for no reason.

In this respect I follow Wald a little better because he explicitly uses projection operators $h^a_b$ instead of assuming a-priori the slicing of space-time that we have defined (so he wouldn't express the 3-D projection of 4-D objects in terms of lapse and shift, but simply in terms of this projection operator). What confuses me about him is that he seems to use this operator to project both covariant and contravariant indices. It makes sense to me to project contravariant indices because $T\Sigma$ (where $\Sigma$ is the hypersurface) is a vector subspace of TM (where M is our manifold), but it doesn't make sense to me to use this operator to project the covariant indices because $T^*\Sigma$ is not a vector subspace of T*M. In fact, one forms which live on T*M naturally induce one forms which live on $T^*\Sigma$ by simply limiting their action. Is he using the projection operator for covariant indices to simply mean this limiting of their action?

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jambaugh
Gold Member
Breaking out my copy of MTW...

Uhggg. Not very clear at all. Note that the 4-coordinates uses are such that the fourth is constant along the embedded 3-surface. Thus the act of projecting a 4-vector expanded in the coordinate basis to one which is tangent to the 3-surface is simply to set the fourth (normal) component to zero. When using this in sums one would thus simply not sum over this component. One can project it to 0 and sum 0 or just not bother summing over that one index value.

This of course only has meaning when we pick coordinates in the embedded 3-surface plus one orthogonal to it. Note also that this is preliminary to extrinsic curvature. We do not want to use this scheme for 3-surface parallel transport as we can (in the analogy of the equatorial vector pointing at the north star) 4-parallel transport the vector along the surface until it has zero tangential component (like the north pointing vector transported to the north pole).

MTW is big and comprehensive but I do not find it at all clarifying nor a good text on the math nor the differential geometry.

As to Wald's exposition I don't have a copy. But I believe your supposition is correct. A better way to think of it is to compose the projection operator with the dual vectors (which is it's dual action). This is the "naturally induced"..."limiting" of "their action".

The resulting dual tensor do technically live in the T*M space but do project out all but TΣ components of TM. They are thus isomorphic to elements of T*Σ and I believe the author is implicitly identifying them via this isomorphism mapping.

Gold Member
The problem I have with the first paragraph above is that MTW in that section is explicitly not using orthonormal coordinate basis vectors. At least, if they are, they pulled a fast one on me because everything before that has $\mathbf{e}_0$ not being a normal vector (one can see several pages before hand that the metric has time-space components). As such, it doesn't seem right to be able to simply not sum over the 0th coordinate.

you can also look at Poisson's book, where I seem to remember he does 3+1 systematically and carefully. (Wald is careful, but the organization is a bit weird to my taste.)

Poisson will always be explicit about "limiting their action" (defining objects with "mixed" spacetime and spatial indices to do precisely that), whereas Wald likes to use abstract spacetime indices everywhere and just regard certain tensors as being "3D".

jambaugh
Gold Member
The problem I have with the first paragraph above is that MTW in that section is explicitly not using orthonormal coordinate basis vectors. At least, if they are, they pulled a fast one on me because everything before that has $\mathbf{e}_0$ not being a normal vector (one can see several pages before hand that the metric has time-space components). As such, it doesn't seem right to be able to simply not sum over the 0th coordinate.

My eyes cross as I try to parse their reasoning. I think the best way to understand this is to a.) work in a similar example of a 2-surface embedded in a 3-manifold, and to work out an explicit example. It may be tedious but it should give some insight. Try e.g. using Euclidean 3-space as one's manifold but use non-orthogonal coordinates, say hyperbolic coordinates, x = r cosh s, y = r sinh s, z = z.

But aside from that let's think about it. From the exposition I think it is clear MTW is referring to the hypersufrace defined by a the fourth coordinate = constant.

Let me refresh my recollection of covariant vs. contravariant bases....
Hmmm....

The gradient of a coordinate (a function of the points on the manifold yielding that coordinate) defines the corresponding coordinate dual basis vector. These, (given a metric), are orthogonal to the surfaces of constant coordinate value but in truth the gradients of scalars live in the cotangent space and we use the metric to transpose them to tangent vectors. A better statement is that the gradient of a scalar function is a dual vector which is co-tangent to the tangent subspace of the sub-manifold defined by setting that function to a constant.

The coordinate basis vectors are parallel to the coordinate curves (curves where all other coordinates are constant).

As such the basis and dual basis are "co-orthogonal" in that (even absent a metric) dual basis vectors for one coordinate will contract with basis vectors for the other coordinates to yield zero.

We can thus project onto a tangent subspace defined tangent to the constraint surface $\Sigma$ defined by the coordinate constraint $x^0=const.$ by a.) setting the 0th component of $a^\mu$ to zero. The tangent basis vector $\mathbf{e}_0$ will not be normal to this tangent subspace but its dual $\boldsymbol{\omega}^0$ will be "co-orthogonal". (And without metric information normality ceases to be defined buto "co-orthogonality" is always meaningful $\boldsymbol{\omega}[\mathbf{e}] = 0$.)

I think the answer to your original inquiry about MTW is that the dropping of the fourth index from the sum is the more proper procedure as it gives the same result without reference to the metric. It is how we project tangent vectors onto the tangent subspace even if no metric is defined. Our usual practice of using a normal vector is metric dependent (but one is using the metric and dual metric twice, ((once to define the normal vector as transpose of "co-orthogonal" dual and once to use it on other tangent vectors)) in a way that cancels out). Projection is independent of the metric.

Gold Member
But there's no way to project a vector in TM to a vector in TS (where S is the sub-manifold) without a metric. Otherwise, your projection will be coordinate dependent which would not be a very good projection.

Gold Member
I figured out the problem. The (3) Christoffel symbols are NOT the same as the (4) Christoffel symbols restricted to i,j,k indices (as I had assumed, since one sees a similar expression towards the end of the same page in MTW). One must define them in a correct way.

...

Ben Niehoff
Gold Member
But there's no way to project a vector in TM to a vector in TS (where S is the sub-manifold) without a metric. Otherwise, your projection will be coordinate dependent which would not be a very good projection.

True.

However, if we have a connection on M, we can pull it back to a connection on S without reference to any metric or projections. MTW are probably restricting their discussion to Levi-Civita connections, where of course you always have a metric (in fact, to discuss things like "spacelike hypersurfaces" as needed in the ADM formalism, you need some sort of additional structure, for which the metric suffices).

To pull back a connection, you just use the standard coordinate transformation law of the Christoffel symbols, using the embedding maps $x^\mu = x^\mu(y^a)$, where $y^a$ are coordinates on the submanifold (i.e., some parametrization of the surface). Try this for the 2-sphere embedded in R^3; it's a much faster way to get the connection than computing it intrinsically within the 2-sphere.

If you have an implicitly-defined surface, such as $x^2 + y^2 + z^2 = 1$, then things become trickier. But in principle, you should be able to pull back the connection without having to refer to a metric; because connections and their pullbacks are things that exist independently of the metric.

Gold Member
I'm trying to fix my Christoffel symbols so that they work. A weird thing that happens is that because I have to project vectors while I can naturally limit the action of one forms, the Christoffel symbols that appear for the covariant derivative of a vector is different than those that appear for the covariant derivative of a one form...

In essence, I have an extra term proportional to N for the vector derivative...

This can't be right...

EDIT: Again I figured out my problem. What is dt restricted to acting on vectors in the hypersurface? I would have thought that it'd simply be 0, but using the "projection operator" defined by Wald I get dt+n/N where n is the one form dual to the normal vector. This second expression also gives me 0 whenever I act it on any vector normal to the surface.

...

Wait a second, that second thing I put is identically 0. Arg...now I'm back to the Christoffel symbols not matching up again...

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Gold Member
Oh my gosh, those sneaky sneaky bastards! I realize now why they all of a sudden switched to Christoffel symbols of the first kind in the middle of calculations...because that is the one that has the 3-D and 4-D versions matching due to the cancelation one gets when lowering the index with the 4-D metric.

Consequently, one can use entirely the 3-D Christoffel symbols for the covariant derivative of a one form if one switches to these Christoffel symbols of the first kind.

That is sneaky!!!

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Gold Member
So...apparently my last post is also wrong. I don't know, I can't seem to catch a break in this. It seems I'm just not doing this right or there's something really funky going on...

Because of the projection, and because $\pi(\mathbf{e}_0)=\mathbf{N}$ where π is the projection operator and N is the shift vector, I get that the Christoffel symbols I must use for the covariant derivative of a vector field are $\Gamma'^i_{jk}=\Gamma^i_{jk}+N^i\Gamma^0_{jk}$ where the ' refers to the fact that the left hand side is the 3-dimensional Christoffel symbols.

Because I don't have to project anything when taking the covariant derivative of a one form, since I can just get rid of dx^0 in that expression, I don't have the funky $N^i$ in there...and so the two expressions don't match.

I.e. I have to use 2 different sets of Christoffel symbols for the 2 slightly different derivatives.

I also have absolutely no idea how to interpret something like $N^i\Gamma^0_{jk}$. What the heck is that?

I can't find an error in the term with the N though, and it seems that that MUST be the correct term because going at this from another perspective: $$\nabla_i \mathbf{e}_j=\Gamma^\mu_{ij}\mathbf{e}_\mu$$ I get: $$\nabla_i \mathbf{e}_j=\Gamma^0_{ij}(N\mathbf{n}+\mathbf{N})+\Gamma^k_{ij}\mathbf{e}_k$$ Now if I want to project away the normal component to get the derivative inside the hypersurface, and I demand that $$\nabla'_i\mathbf{e}_j=\Gamma'^k_{ij}\mathbf{e}_k$$ then I must have the same expression for Gamma' I had earlier.

I arrived at this same conclusion from 2 different ways, so unless I'm just doing something completely wrong...I think that at least that expression is correct.

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...A better way to think of it is to compose the projection operator with the dual vectors (which is it's dual action). This is the "naturally induced"..."limiting" of "their action".

The resulting dual tensor do technically live in the T*M space but do project out all but TΣ components of TM. They are thus isomorphic to elements of T*Σ and I believe the author is implicitly identifying them via this isomorphism mapping...

jimbaugh, reading your response here I could not help but be reminded of the dual space relationship between any two observers moving relative to each other. For any two moving observers you can always find a rest system in which the two observers are moving in opposite directions at the same speed. When you draw the space-time diagram you see that each of the mover's coordinates is the dual of the other. I've sketched it below. 